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\begin{center}
\vskip 1cm{\LARGE\bf
Combinatorial Proofs of Some Stirling Number Convolution Formulas\\
}
\vskip 1cm
\large
Mark Shattuck\\
Department of Mathematics\\
University of Tennessee\\
Knoxville, TN\\
USA\\
\href{mailto:mshattuc@utk.edu}{\tt mshattuc@utk.edu}\\
\end{center}
\vskip .2 in
\begin{abstract}
Recently, some new convolution formulas extending the orthogonality
of the Stirling numbers of the first and second kind were shown by
algebraic techniques. The formulas involve sums of products of the two
Stirling numbers wherein the inner arguments vary while differing by a
prescribed amount and the outer arguments are fixed. Here, we provide
combinatorial proofs of these formulas using direct enumeration and
sign-changing involutions. Our arguments may be extended to establish
generalizations of the foregoing results in terms of the $r$-Stirling
numbers.
\end{abstract}
\section{Introduction}
Perhaps two of the most prevalent number sequences in enumerative
combinatorics are the Stirling numbers of the first and second kind,
which will be denoted here by $\stirling{n}{k}$ and $\Stirling{n}{k}$
respectively in accordance with \cite{Kn}. See sequences \seqnum{A008275} and
\seqnum{A008277} in the {\it On-Line Encyclopedia of Integer Sequences}
\cite{Slo} and references contained therein. Recall
that $\stirling{n}{k}$ counts the permutations of $[n]=\{1,\ldots,n\}$
having $k$ cycles (and is often referred to as the \emph{signless}
Stirling number of the first kind), whereas $\Stirling{n}{k}$ enumerates
the partitions of $[n]$ with $k$ blocks. Both kinds of Stirling numbers
satisfy a variety of identities; see, for example, the texts \cite[Chap.\
V]{Com}, \cite[\S6.1]{Gr}, and \cite{Qu} as well as the recent papers
\cite{Cai,Chu,Guo,Sh}.
We will need the following further notation. Define the generalized harmonic number $\sigma_{\lambda}(m;n)$ by
$$\sigma_{\lambda}(m;n)=\sum_{m \leq i_1<\cdots< i_\lambda \leq n}\prod_{j=1}^\lambda \frac{1}{i_j}, \qquad \lambda>0,$$
with $\sigma_0(m;n)=1$. Denote the Lah number (see \cite[\seqnum{A008297}]{Slo}) by $\Lah{n}{k}$, which counts the partitions of $[n]$ into $k$ \emph{contents-ordered} blocks (i.e., blocks in which the order of the elements contained therein matters). Recall that $\Lah{n}{k}$ is given explicitly by $\Lah{n}{k}=\frac{n!}{k!}\binom{n-1}{k-1}$ for $1 \leq k \leq n$, which is also seen to hold for $k=0$ and all $n \geq 0$ if one adopts the convention $\binom{i}{-1}=\delta_{i,-1}$ for integers $i \geq -1$.
Chu \cite{Chu} considered the following general sums where $\lambda$ denotes an arbitrary integer:
\begin{equation}\label{Phi}
\Phi_{m,n}(\lambda)=\sum_{k=m}^{n-\lambda}\stirling{n}{k+\lambda}\Stirling{k}{m}
\end{equation}
and
\begin{equation}\label{Psi}
\Psi_{m,n}(\lambda)=\sum_{k=m}^{n-\lambda}\stirling{n}{k+\lambda}\Stirling{k}{m}(-1)^{n-k}.
\end{equation}
Note that the evaluations of \eqref{Phi} and \eqref{Psi} when $\lambda=0$ correspond to the well-known identities
$$\sum_{k=m}^n\stirling{n}{k}\Stirling{k}{m}=\Lah{n}{m}$$
and
$$\sum_{k=m}^n\stirling{n}{k}\Stirling{k}{m}(-1)^{n-k}=\delta_{m,n},$$
see, e.g., \cite[\S6.1]{Gr}.
Further, the evaluation of \eqref{Psi} when $\lambda=1$ is known and gives
$$\sum_{k=m}^{n-1}\stirling{n}{k+1}\Stirling{k}{m}(-1)^{n-k}=(-1)^{n-m}\frac{(n-1)!}{m!},$$
which corresponds to \cite[Identity~6.25]{Gr}.
The following general identities for $\Phi_{m,n}(\lambda)$ and $\Psi_{m,n}(\lambda)$ were established in \cite{Chu} by algebraic methods using a connection coefficient approach:
\begin{align}
\Phi_{m,n}(\lambda)&=\sum_{j=m}^{n-\lambda}\frac{(n-1)!}{m!}\binom{j-1}{m-1}\sigma_{\lambda-1}(j+1;n-1), \quad \lambda>0,\label{Phipos}\\
\Phi_{m,n}(\lambda)&=\sum_{i=0}^{-\lambda}\sum_{j=0}^i\frac{(-1)^{\lambda+j}}{n^{\lambda+i}}\binom{-\lambda}{i}\Stirling{i}{j}\Lah{j+n}{m}, \quad \lambda<0, \label{Phineg}
\end{align}
\begin{align}
\Psi_{m,n}(\lambda)&=(-1)^{n-m}\frac{(n-1)!}{m!}\sigma_{\lambda-1}(m+1;n-1), \quad \lambda>0, \label{Psipos}\\
\Psi_{m,n}(\lambda)&=\sum_{i=0}^{-\lambda}\frac{(-1)^\lambda}{n^{\lambda+i}}\binom{-\lambda}{i}\Stirling{i}{m-n}, \quad \lambda<0. \label{Psineg}
\end{align}
Throughout, one may assume $n,m \geq 0$ with $\lambda \in \mathbb{Z}$, where it is seen that the lower index of summation in the definitions of $\Phi_{m,n}(\lambda)$ and $\Psi_{m,n}(\lambda)$ may be replaced by $k=\max\{m,-\lambda\}$. Note also that both $\Phi_{m,n}(\lambda)$ and $\Psi_{m,n}(\lambda)$ are zero if $n0$ and $m \leq k \leq n-\lambda$, first arrange the members of $[n]$ according to a permutation having $k+\lambda$ cycles $C_1,\ldots,C_{k+\lambda}$, where the smallest element is first in each cycle and cycles are ordered according to the size of their respective first elements. We then arrange the first $k$ cycles $C_1,\ldots,C_k$ according to a partition having $m$ blocks, where cycles within a block are written in decreasing order of their first elements. Note that one may then erase the pair of parentheses enclosing each cycle so that now cycle starters correspond to left-right minima within $m$ contents-ordered blocks. The remaining $\lambda$ cycles $C_{k+1},\ldots,C_{k+\lambda}$ are set aside. Let $\mathcal{A}_k=\mathcal{A}_k^{(m,n,\lambda)}$ denote the set of configurations of $[n]$ that arise in this manner for $m \leq k \leq n-\lambda$. Define $\mathcal{A}=\mathcal{A}^{(m,n,\lambda)}$ by $\mathcal{A}=\cup_{k=m}^{n-\lambda}\mathcal{A}_k$. Then it is seen that $\Phi_m(\lambda)$ gives the cardinality of $\mathcal{A}$.
We now enumerate members of $\mathcal{A}$ by considering the smallest element $j+1$ belonging to cycle $C_{k+1}$. To do so, we arrange the elements of $[j]$ according to an arbitrary permutation (written using the one-line notation) and insert $m-1$ internal dividers in any of $\binom{j-1}{m-1}$ ways. Then elements between consecutive dividers (or preceding the first or following the last divider) constitute the elements within one of $m$ (nonempty) contents-ordered blocks. Since the ordering of the blocks themselves is immaterial, we divide by $m!$, and hence there are $\frac{j!}{m!}\binom{j-1}{m-1}=\Lah{j}{m}$ ways in which to arrange the members of $[j]$. Further, the number of ways in which to insert the elements of $[j+2,n]$ so that exactly $\lambda-1$ additional cycles are created is given by
$$\frac{(n-1)!}{j!}\sum_{j+1\leq i_1<\cdots1$ (which also holds if $\lambda=1$ since $\sigma_0(j+1;n-1)=1$). To see this, note that $i_1+1,\ldots,i_{\lambda-1}+1$, where $i_\ell$ is as in the preceding multi-sum, would correspond to the starters of cycles $C_{k+2},\ldots,C_{k+\lambda}$, respectively. Hence, each factor $i_\ell$ for $1 \leq \ell \leq \lambda-1$ is ``missed'' in the product $(j+1)(j+2)\cdots(n-1)=\frac{(n-1)!}{j!}$. Thus, for each $j \in [m,n-\lambda]$, there are $\Lah{j}{m}\cdot \frac{(n-1)!}{j!}\sigma_{\lambda-1}(j+1;n-1)=\frac{(n-1)!}{m!}\binom{j-1}{m-1}\sigma_{\lambda-1}(j+1;n-1)$ members of $\mathcal{A}_j$ and considering all possible $j$ completes the proof. \end{proof}
\begin{proof}[Proof of \eqref{Phineg}] Let $n,m \geq 0$ be fixed with $\lambda<0$. Given $0 \leq j \leq i \leq -\lambda$, let $\mathcal{B}_{i,j}$ denote the set of configurations obtained as follows. Choose exactly $i$ of the elements of $I=[n+1,n-\lambda]$ and arrange them according to an ordinary partition with $j$ blocks. We then arrange these $j$ blocks $B_1,\ldots,B_j$, together with the members of $[n]$, according to a Lah distribution having $m$ blocks. Then insert the $-\lambda-i$ unchosen elements of $I$ into this Lah distribution so that at the time of insertion each element is placed so that it directly precedes some member of $[n]$ (and not any of the blocks $B_\ell$ for $\ell \in [j]$). Further, if $S=\{s_1>\cdots>s_{-\lambda-i}\}$ denotes the set of elements to be inserted, then $s_1$ is to be added first directly prior to some member of $[n]$ within one of the $m$ blocks. The element $s_2$ is to be added in the same way, where if $s_2$ is placed in the same slot as $s_1$ was, then $s_2$ is to follow $s_1$ (this ensures that $s_2$ is also inserted directly prior to some element of $[n]$). The remaining elements $s_3,\ldots,s_{-\lambda-i}$ are then to be added sequentially in this same manner. Then the absolute value of the generic summand in the formula on the right-hand side of \eqref{Phineg} is seen to give $|\mathcal{B}_{i,j}|$ for all $i$ and $j$. Define the sign of a member of $\mathcal{B}_{i,j}$ to be $(-1)^{\lambda+j}$ and let $\mathcal{B}=\cup_{i=0}^{-\lambda}\cup_{j=0}^i\mathcal{B}_{i,j}$. Then the right side of \eqref{Phineg} gives the sum of the signs of all members of $\mathcal{B}$.
To complete the proof, we define a sign-changing involution on $\mathcal{B}$. To do so, we will need the following further definitions. First, we will refer to the blocks of $\pi \in \mathcal{B}$ whose members themselves are either individual elements of $[n-\lambda]$ or blocks whose elements belong to $I$ as \emph{super-blocks}. We will refer to an internal block consisting of members of $I$ within a super-block as a \emph{sub-block}. Note that each member of $\mathcal{B}_{i,j}$ contains exactly $j$ sub-blocks in all and each member of $\mathcal{B}$ has exactly $m$ super-blocks, the contents of which we now describe in further detail. By a \emph{free} element within a super-block $B$, we mean a member of $I$ not belonging to any of the sub-blocks lying within $B$. Given $a \in [n]$, let $\mathcal{L}_a$ denote the (ordered) set consisting of free elements and/or sub-blocks occurring to the left of $a$ if $a$ is the leftmost member of $[n]$ lying within its super-block or between $a$ and the next member of $[n]$ to its left if not. If $b$ is the rightmost member of $[n]$ in its super-block, then let $\mathcal{M}_b$ denote the collection of sub-blocks occurring to the right of $b$. Note that since free elements are to precede directly elements of $[n]$, no free elements can occur to the right of $b$ since it is rightmost and thus $\mathcal{M}_b$ consists solely of sub-blocks. Finally, let $\mathcal{N}_c$ for $c \in I$ denote the contents of a super-block of $\pi$ that contains no elements of $[n]$ and whose smallest element lying within one of the sub-blocks is $c$. Note that $\mathcal{N}_c$, like $\mathcal{M}_b$, can contain no free elements. Further, within a nonempty $\mathcal{L}_a$, the sub-blocks are followed by any free elements (in decreasing order). Given $b \in [n]$ or $c \in I$, let $\mathcal{M}_b$ be empty if $b$ is not the rightmost element of $[n]$ in its super-block and let $\mathcal{N}_c$ be empty if $c$ does not correspond to the smallest element lying within a sub-block of a super-block containing no elements of $[n]$.
Let $p$ denote the smallest index $j \in [n-\lambda]$ such that at least one of the following holds:
\begin{align*}
(i)&~j \in [n], \text{ with } \mathcal{L}_j \text { nonempty},\\
(ii)&~j\in [n] \text{ and } \mathcal{M}_j \text{ is nonempty, with } \mathcal{M}_j\neq \{a_1\},\ldots,\{a_k\} \text{ for some } k \geq 1,\\
(iii)&~j\in I \text{ and } \mathcal{N}_j \text{ is nonempty, with } \mathcal{N}_j\neq \{a_1\},\ldots,\{a_k\} \text{ for some } k \geq 1,
\end{align*}
where $a_1<\cdots\cdots>b_s$ and $s \geq 1$, then we change the status of $m=\max\{a_1,\ldots,a_r,b_1,\ldots,b_s\}$. Note that $m$ occurs either as $K_r=\{m\}$ or as $b_1=m$; thus, we may either erase the brackets enclosing $m$ and designate $m$ a free element if the former, or vice versa, if the latter. This operation again reverses the sign and completes the pairing of all possible $\pi$ for which $\mathcal{L}_i$ is nonempty for some $i$.
Let $\mathcal{B}^*\subseteq \mathcal{B}$ denote the set of survivors of the involution above. Then $\mathcal{B}^*$ comprises those configurations in which $\mathcal{L}_i$ is empty for all $i \in [n]$, with $\mathcal{M}_j$ and $\mathcal{N}_j$ consisting of singletons (possibly none) arranged in ascending order for all $j$. Then each member of $\mathcal{B}^*$ has positive sign and $|\mathcal{B}^*|=\sum_{k=m}^{n-\lambda}\stirling{n}{k+\lambda}\Stirling{k}{m}$. To realize this, first note that any left-right minima belonging to $[n]$ within a super-block of $\rho \in \mathcal{B}^*$ may be viewed as a cycle starter (in some permutation of $[n]$). If we denote the number of these cycles by $k+\lambda$, then it is seen that there are $k$ objects altogether ($k+\lambda$ cycles and $-\lambda$ singleton sub-blocks) to be arranged in the $m$ super-blocks of $\rho$, wherein the members of $I$ (written as singleton blocks) within a super-block must occur in ascending order following any elements of $[n]$ which are arranged in cycles written in decreasing order. Thus, given any permutation of $[n]$ with $k+\lambda$ cycles, there are $\Stirling{k}{m}$ ways in which to arrange the cycles as described together with the $-\lambda$ elements of $I$ so as to form a member of $\mathcal{B}^*$. Considering all possible $k$ implies $|\mathcal{B}^*|$ is as stated and completes the proof. \end{proof}
\begin{proof}[Proof of \eqref{Psipos}] Given $\lambda>0$ and $m \leq k \leq n-\lambda$, let $\mathcal{C}_k=\mathcal{C}_k^{(m,n,\lambda)}$ denote the set of configurations that are obtained by first arranging the elements of $[n]$ according to a permutation having $k+\lambda$ cycles expressed in standard form and then forming a partition with $m$ blocks using the first $k$ of these cycles. Here, cycles are assumed to be ordered by the size of their respective smallest (= first) elements. Define the sign of each member of $\mathcal{C}_k$ to be $(-1)^{n-k}$. Let $\mathcal{C}=\mathcal{C}^{(m,n,\lambda)}$ be given by $\mathcal{C}=\cup_{k=m}^{n-\lambda}\mathcal{C}_k$. Then $\Psi_{m,n}(\lambda)$ gives the sum of the signs of all members of $\mathcal{C}$.
We define an involution on $\mathcal{C}$ as follows. Given $\pi \in \mathcal{C}$, identify the block $B$ of $\pi$ containing the smallest element of $[n]$ within its cycles out of all the blocks of $\pi$ that contain at least two elements of $[\ell]$ within their respective cycles, where $\ell+1$ denotes the first element of the $\lambda$-th largest cycle of $\pi$ (i.e., if $\pi \in \mathcal{C}_k$ for some $k$, then $\ell+1$ is the first element of the $(k+1)$-st cycle). Let $a$ and $b$ where $a**n$, first note that all cycles must again have length one with these cycles going in $n$ distinct blocks of the partition within a survivor of the involution. Then we select $\binom{-\lambda}{i}$ members of $[n+1,n-\lambda]$ to comprise $m-n$ additional blocks, which can be effected in $\Stirling{i}{m-n}$ ways. Finally, the remaining members of $I$ go in the blocks already containing the $1$-cycles, which can be achieved in $n^{-\lambda-i}=\frac{1}{n^{\lambda+i}}$ ways. Considering all possible $i$, where $0 \leq i \leq -\lambda$, implies the formula in the case $m>n$ and completes the proof of \eqref{Psineg}. \end{proof}
We conclude this section by providing combinatorial proofs of some related recurrences for $\Phi_{m,n}(\lambda)$ and $\Psi_{m,n}(\lambda)$ which were shown in \cite{Chu} by other methods. Counting members of the set $\mathcal{A}$ from the proof of \eqref{Phipos} in another way yields the following recurrence for $\Phi_{m,n}(\lambda)$:
\begin{equation}\label{Phirec}
\Phi_{m,n}(\lambda)=\sum_{i=m+\lambda-1}^{n-1}\frac{(n-1)!}{i!}\Phi_{m,i}(\lambda-1), \quad \lambda>0.
\end{equation}
\begin{proof}
One may assume $n\geq m+\lambda$, for otherwise both sides of \eqref{Phirec} are zero. Consider the smallest element $i+1$ of the last cycle within a member of $\mathcal{A}^{(m,n,\lambda)}$, where $m+\lambda-1 \leq i \leq n-1$. Then the elements of $[i]$ are arranged according to some member of $\mathcal{A}^{(m,i,\lambda-1)}$ with $i+1$ going in a new cycle by itself. Elements of $[i+2,n]$ are then inserted sequentially so that each $x \in [i+2,n]$ directly follows some element of $[x-1]$ occurring in either one of the $m$ (contents-ordered) blocks or in one of the $\lambda$ cycles that are set aside. Note that no $x$ can be placed at the very beginning of one of the $m$ blocks since this would create implicitly a new cycle with first element $x$ contradicting that $i+1$ was the smallest element of the last cycle. Thus, there are $(i+1)(i+2)\cdots (n-1)=\frac{(n-1)!}{i!}$ ways in which to insert the elements of $[i+2,n]$ and considering all possible $i$ implies \eqref{Phirec}.
\end{proof}
The argument used to show \eqref{Psipos} can be extended to prove the following recurrence relations from \cite{Chu}
for $\Psi_{m,n}(\lambda)$ where $\lambda>0$:
\begin{align}
\Psi_{m,n}(\lambda)&=\sum_{i=m+1}^{n-\lambda+1}(-1)^{i-m}\frac{(i-1)!}{m!}\Psi_{i,n}(\lambda-1)\label{Psirec1}\\
&=\sum_{i=m+\lambda-1}^{n-1}(-1)^{n-i}\frac{(n-1)!}{i!}\Psi_{m,i}(\lambda-1)\label{Psirec2}.
\end{align}
\begin{proof}
Note that one may assume $\lambda>1$ in \eqref{Psirec1} and \eqref{Psirec2}, as they are seen to hold when $\lambda=1$ since $\Psi_{m,n}(1)=(-1)^{n-m}\frac{(n-1)!}{m!}$ and $\Psi_{i,n}(0)=\delta_{i,n}$. By the proof of \eqref{Psipos}, we have $\Psi_{m,n}(\lambda)=(-1)^{n-m}\stirling{n-m-1}{\lambda-1}_{m+1}$. Upon replacing $m$ by $m-1$ and $\lambda$ by $\lambda+1$ in \eqref{Psirec1} and \eqref{Psirec2}, one may then show equivalently
\begin{align}
\stirling{n-m}{\lambda}_m&=\sum_{i=m}^{n-\lambda}\frac{(i-1)!}{(m-1)!}\stirling{n-i-1}{\lambda-1}_{i+1} \label{Psirec1a}\\
&=\sum_{i=m+\lambda-1}^{n-1}\frac{(n-1)!}{i!}\stirling{i-m}{\lambda-1}_m.\label{Psirec2a}
\end{align}
Before proceeding with the proof of \eqref{Psirec1a}, let us introduce a couple of definitions which we will also make use of in the next section. Permutations enumerated by the $r$-Stirling number $\stirling{n}{k}_r$ are called $r$-permutations and we will refer to a cycle containing a member of $[r]$ within an $r$-permutation as \emph{special}, with all other cycles being \emph{non-special}. The same descriptors will also be used for the elements themselves of the sets $[r]$ and of $[r+1,n+r]$, respectively. Analogous terminology can be applied to the $r$-partitions enumerated by $\Stirling{n}{k}_r$. Assume that the smallest element is first in all cycles. To show \eqref{Psirec1a}, we argue that the right side counts the permutations enumerated by $\stirling{n-m}{\lambda}_m$ according to the smallest element $i+1$ of the first non-special cycle where $m \leq i \leq n-\lambda$. Note that there are $\frac{(i-1)!}{(m-1)!}$ choices regarding the placement of the elements of $[i]$ since the elements of $[m]$ belong to different cycles, with $i+1$ going in a new cycle by itself. At this point, we may regard each of the $i+1$ elements already inserted as special (each starting its own respective cycle) when placing the elements of $[i+2,n]$. Note that $\lambda-1$ additional non-special cycles must be created when these elements are placed since already one non-special cycle has been started. Thus, there are $\stirling{n-i-1}{\lambda-1}_{i+1}$ possibilities concerning the placement of the elements of $[i+2,n]$ and considering all $i$ completes the proof of \eqref{Psirec1a}. A similar proof applies to \eqref{Psirec2a}, where instead one considers the element $i+1$ starting the \emph{last} non-special cycle for some $i\in[m+\lambda-1,n-1]$. \end{proof}
\section{Generalization}
Given $r, s \geq 0$, let
\begin{equation}\label{genPhi}
\Phi_{m,n}^{(r,s)}(\lambda)=\sum_{k=m}^{n-\lambda}\stirling{n}{k+\lambda}_r\Stirling{k}{m}_s
\end{equation}
and
\begin{equation}\label{genPsi}
\Psi_{m,n}^{(r,s)}(\lambda)=\sum_{k=m}^{n-\lambda}\stirling{n}{k+\lambda}_r\Stirling{k}{m}_s(-1)^{n-k}.
\end{equation}
In this section, we find formulas for \eqref{genPhi} and \eqref{genPsi} which generalize those above corresponding to the case $r=s=0$. Let $x^{\overline{m}}=x(x+1)\cdots(x+m-1)$ for a positive integer $m$, with $x^{\overline{0}}=1$. We first consider $\Phi_{m,n}^{(r,s)}(\lambda)$.
\begin{theorem}\label{Phigenth}
We have
\begin{equation}\label{Phigenpos}
\Phi_{m,n}^{(r,s)}(\lambda)=\sum_{\ell=m+\lambda}^n\sum_{j=m}^{\ell-\lambda}\frac{(\ell-1)!r^{\overline{n-\ell}}}{m!}\binom{n}{\ell}\binom{j+s-1}{m+s-1}\sigma_{\lambda-1}(j+1;\ell-1), \quad \lambda>0,
\end{equation}
and
\begin{equation}\label{Phigenneg}
\Phi_{m,n}^{(r,s)}(\lambda)=\sum_{\ell=0}^n\sum_{i=0}^{-\lambda}\sum_{j=0}^i\frac{(-1)^{\lambda+j}(\ell+j)!r^{\overline{n-\ell}}}{\ell^{\lambda+i}m!}\binom{n}{\ell}\binom{-\lambda}{i}\Stirling{i}{j}\binom{\ell+j+s-1}{m+s-1}, \quad \lambda<0.
\end{equation}
\end{theorem}
\begin{proof}
To show \eqref{Phigenpos}, we first generalize the sets $\mathcal{A}_k$ and $\mathcal{A}$ from the proof of \eqref{Phipos} above. Define $\mathcal{A}_k^{(r,s)}$ to be the extension of $\mathcal{A}_k$ obtained by replacing $[n]$ with $[n+r]$ and allowing some of the elements of $J=[r+1,n+r]$ to go in special cycles starting with elements of $[r]$ and then allowing some of the first $k$ non-special cycles to go in $s$ special blocks when forming a partition of cycles according the some $s$-partition enumerated by $\Stirling{k}{m}_s$. Let $\mathcal{A}^{(r,s)}=\cup_{k=m}^{n-\lambda}\mathcal{A}_k^{(r,s)}$, which extends the prior set $\mathcal{A}$. Note that $\Phi_{m,n}^{(r,s)}$ gives $|\mathcal{A}^{(r,s)}|$. To count the members of $\mathcal{A}^{(r,s)}$ in another way, first suppose $n-\ell$ elements of $J$ are to go in the special cycles. Then there are $\binom{n}{\ell}r^{\overline{n-\ell}}$ ways in which to choose and arrange these elements. Let $U=\{u_1<\cdots0$ such as
$$\sum_{k=m}^{n-\lambda}\stirling{n}{k+\lambda}\Stirling{k}{m}_r=\sum_{j=m}^{n-\lambda}\frac{(n-1)!}{m!}\binom{j+r-1}{m+r-1}\sigma_{\lambda-1}(j+1;n-1)$$
and
$$\sum_{k=m}^{n-\lambda}\stirling{n}{k+\lambda}_r\Stirling{k}{m}=\sum_{\ell=m+\lambda}^n\sum_{j=m}^{\ell-\lambda}\frac{(\ell-1)!r^{\overline{n-\ell}}}{m!}\binom{n}{\ell}\binom{j-1}{m-1}\sigma_{\lambda-1}(j+1;\ell-1).$$
Note that letting $r=0$ in either of these formulas recovers \eqref{Phipos}. Taking $\lambda=-1$ in \eqref{Phigenneg} yields
$$\sum_{k=m}^{n+1}\stirling{n}{k-1}_r\Stirling{k}{m}_s=\sum_{\ell=0}^n\frac{\ell!r^{\overline{n-\ell}}}{m!}\binom{n}{\ell}\left((\ell+1)\binom{\ell+s}{m+s-1}-\ell\binom{\ell+s-1}{m+s-1}\right).$$
There are the following comparable formulas for $\Psi_{m,n}^{(r,s)}(\lambda)$.
\begin{theorem}\label{Psigenth}
We have
\begin{equation}\label{Psigenpos}
\Psi_{m,n}^{(r,s)}(\lambda)=\sum_{i=0}^u\sum_{j=0}^{u-i}\frac{(-1)^{n-m-j}(n-i-1)!r^{\overline{i}}}{m!}\binom{n}{i}\binom{s}{j}\sigma_{\lambda-1}(m+j+1;n-i-1), \quad \lambda>0,
\end{equation}
and
\begin{equation}\label{Psigenneg}
\Psi_{m,n}^{(r,s)}(\lambda)=\sum_{i=0}^n\sum_{j=0}^{n-i}\sum_{\ell=0}^{-\lambda}\frac{(-1)^{\lambda+i}j!r^{\overline{i}}}{(v_{i,j}+s)^{\lambda+\ell}}\binom{n}{i}\binom{s}{j}\binom{n-i}{j}\binom{-\lambda}{\ell}\Stirling{\ell}{m-v_{i,j}}, \quad \lambda<0,
\end{equation}
where $u:=n-m-\lambda$ and $v_{i,j}:=n-i-j$.
\end{theorem}
\begin{proof}
To show \eqref{Psigenpos}, let $\mathcal{C}_k^{(r,s)}$ be obtained by first forming $r$-permutations of $[n+r]$ having $k+\lambda$ non-special cycles and then arranging the first $k$ non-special cycles according to an $s$-partition having $m$ non-special blocks. Let members of $\mathcal{C}_k^{(r,s)}$ have sign $(-1)^{n-k}$ and $\mathcal{C}^{(r,s)}=\cup_{k=m}^{n-\lambda}\mathcal{C}_k^{(r,s)}$. Then $\Psi_{m,n}^{(r,s)}$ gives the sum of the signs of all members of $\mathcal{C}^{(r,s)}$. Applying the involution used in the proof of \eqref{Psipos} above (considering also any cycles placed in the $s$ special blocks, if necessary) implies that the set $M$ of survivors consists of those members of $\mathcal{C}^{(r,s)}$ in which $k=m+j$ for some $0 \leq j \leq s$ wherein the $k+1$ smallest members of $J$ belonging to non-special cycles each start their own cycle, with the first $k$ of these cycles lying in different blocks of the $s$-partition. If $i$ denotes the number of members of $J$ going in special cycles within a member of $M$, then we have $j \leq u-i$ and there are $\frac{(n-i-1)!}{(m+j)!}\sigma_{\lambda-1}(m+j+1;n-i-1)$ ways in which to arrange those members of $J$ lying in non-special cycles but not starting one of the first $k+1$ non-special cycles. Further, there are $\binom{m+j}{j}\binom{s}{j}j!$ ways to choose and then arrange the cycle starters of the cycles that go in the special blocks within a member of $M$. Upon considering all possible values of $i$ and $j$, the right side of \eqref{Psigenpos} is seen to give the sum of the signs of members of $M$.
To show \eqref{Psigenneg}, we form the set $\mathcal{D}_k^{(r,s)}$ consisting of configurations obtained by first arranging the elements of $[n+r]$ according to an $r$-permutation having $k+\lambda$ non-special cycles and then arranging these non-special cycles, together with the elements of $K$, according to an $s$-partition of size $k+s$ having $m+s$ blocks. Define the sign of members of $\mathcal{D}_k^{(r,s)}$ as $(-1)^{n-k}$ and let $\mathcal{D}^{(r,s)}$ be the union of all $\mathcal{D}_k^{(r,s)}$.
Let $i$ again denote the number of elements of $J$ going in one of the $r$ special cycles and $j$ be the number of special blocks that contain at least one non-special cycle in the $s$-partition. Upon applying the prior involution, to determine the sum of the signs of all members of $\mathcal{D}^{(r,s)}$, we may assume that the non-special cycles all have length one and occur in different blocks of the $s$-partition. Note that if $mv_{i,j}$, then $\ell$ members of $K$ for some $1 \leq \ell \leq -\lambda$ must occupy the remaining unfilled non-special blocks, which implies the remaining terms in the sum account for survivors where $m>v_{i,j}$. Combining this case with the prior cases implies the right side of \eqref{Psigenneg} gives the sum of the signs of all survivors, which completes the proof.
\end{proof}
Letting, for example, $r=0$ or $s=0$ in \eqref{Psigenpos} gives for $\lambda>0$ the identities
$$\sum_{k=m}^{n-\lambda}\stirling{n}{k+\lambda}\Stirling{k}{m}_r(-1)^{n-k}=\sum_{j=0}^u \frac{(-1)^{n-m-j}(n-1)!}{m!}\binom{r}{j}\sigma_{\lambda-1}(m+j+1;n-1)$$
and
$$\sum_{k=m}^{n-\lambda}\stirling{n}{k+\lambda}_r\Stirling{k}{m}(-1)^{n-k}=(-1)^{n-m}\sum_{i=0}^{u}\frac{(n-i-1)!r^{\overline{i}}}{m!}\binom{n}{i}\sigma_{\lambda-1}(m+1;n-i-1).$$
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\noindent 2020 {\it Mathematics Subject Classification}: Primary: 05A19; Secondary: 05A05.
\noindent \emph{Keywords:} combinatorial identity, Stirling number, Lah number, combinatorial proof.
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\noindent (Concerned with sequences \seqnum{A008275}, \seqnum{A008277}, and \seqnum{A008297}.)
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\noindent
Received August 25 2021;
revised version received January 9 2022.
Published in {\it Journal of Integer Sequences}, January 29 2022.
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